Answer:
a. Concave down
Linear increasing
b. Increases the reaction rate
c. The reaction approaches the saturation point of the enzyme
Explanation:
a. For the reaction with enzyme, the shape is concave down. The action of the enzyme on the preferred substrate is initially very rapid and decreases as the enzyme becomes saturated and the ratio of products to substrate increases to approach an equilibrium rate of reaction
For the reaction without enzyme, the shape is linear and increasing. Increase in the concentration of the substrate will increase the number of effective collisions that lead into product formation leading to an increased rate of the chemical reaction
b. The enzyme increases the proportion of effective combination of substrates to form the products
c. The curve of the reaction with enzyme flattens out because as the concentration of the substrate increases while that of the enzyme remains the same, the enzyme becomes saturated and less able to increase the rate of the reaction of the excess substrate.
The answer to the question is D.
Answer:
It will be denatured.
Explanation:
Salivary amylase works best at pH 6.8.
At pH 2.5, its activity will decrease enormously as it becomes denatured by the high acidity of the stomach contents.
The half-life equation is written as:
An = Aoe^-kt
We use this equation for the solution. We do as follows:
5.5 = 176e^-k(165)
k = 0.02
<span>What is the half-life of the goo in minutes?
</span>
0.5 = e^-0.02t
t = 34.66 minutes <----HALF-LIFE
Find a formula for G(t) , the amount of goo remaining at time t.G(t)=?
G(t) = 176e^-0.02t
How many grams of goo will remain after 50 minutes?
G(t) = 176e^-0.02(50) = 64.75 g
Answer:
d. 1.2 × 1024
Explanation:
From the equation of reaction
2H2 + O2= 2H2O
i.e 2mole(4g) of hydrogen requires 1 mole(32g) of oxygen to produce 2mole (2×6.02×10^23 molecules) of H2O= 1.2×20^24 molcules of water.
NB: 1 mole of H2O contains 6.02×10^23 molecules of H2O
